Quantitative Aptitude – Caselets
Program: West Bengal Forest Guard
A caselet is a short, real‑world problem that combines multiple quantitative concepts.
They are frequently used in competitive exams (WBFG, UPSC, etc.) to test your ability to interpret data, model the situation, and solve efficiently.
1. Simple Explanation
| What is a Caselet? | Why it matters? |
|---|---|
| A concise scenario (often 4–6 lines) that contains at least two quantitative concepts (ratio, percentage, time‑work, distance, mixture, profit‑loss, etc.). | It mimics exam questions that require you to chain several calculations in one go, testing both conceptual and speed skills. |
Example of a Caselet
“A forest guard collected 1200 kg of timber in 8 days. The next 5 days he worked 25 % more hours per day. Assuming the same productivity per hour, how many kg of timber will he collect in the next 5 days?”
Notice the need to:
- Find productivity per day.
- Adjust for increased work hours.
- Apply the new productivity to the next 5 days.
2. Key Points to Master Caselets
| Concept | Typical Formula | Quick Trick |
|---|---|---|
| Ratio & Proportion | ( \frac{a:b}{c:d} = \frac{a}{b} = \frac{c}{d} ) | Convert ratios to fractions early. |
| Percentage | ( P% \text{ of } X = \frac{P}{100}\times X ) | Remember “( % ) = per hundred”. |
| Time‑Work | Work = Rate × Time; Rate = Work/Time | Use “Rate per hour” or “Units per day” as a single variable. |
| Distance‑Speed‑Time | ( d = r \times t ) | Express all times in same units (hrs). |
| Mixture Problems | ( \frac{W_1}{W_2} = \frac{C_1}{C_2} ) | Think in terms of “units per litre” or “cost per kg”. |
| Profit & Loss | ( \text{Profit %} = \frac{\text{Profit}}{\text{Cost}}\times100 ) | Convert to “Selling price = Cost × (1 + P%)”. |
| Average | ( \text{Avg} = \frac{\text{Sum of values}}{\text{No. of values}} ) | “Average of two numbers” = (a+b)/2. |
| Increase/Decrease | ( \text{New} = \text{Old}\times(1\pm\frac{I}{100}) ) | Use “+” for increase, “–” for decrease. |
Common Pitfalls
- Unit mismatch – always convert to common units before calculation.
- Misreading “per day” vs “per hour” – keep track of the time unit.
- Ignoring the “same productivity” clause – it often means rate per hour stays constant.
- Skipping intermediate steps – in caselets, small errors stack up.
3. Worked Example
Problem Statement
A forest guard collected 1200 kg of timber in 8 days. The next 5 days he worked 25 % more hours per day. Assuming the same productivity per hour, how many kg of timber will he collect in the next 5 days?
Step‑by‑Step Solution
| Step | What to Do | Calculation | Result |
|---|---|---|---|
| 1 | Find productivity per day (initial). | ( \text{Daily productivity} = \frac{1200\text{ kg}}{8\text{ days}} = 150\text{ kg/day} ) | 150 kg/day |
| 2 | Convert to productivity per hour. | Assume original daily hours = 8 hrs (common assumption). ( \text{Rate per hour} = \frac{150\text{ kg}}{8\text{ hrs}} = 18.75\text{ kg/hr} ) |
18.75 kg/hr |
| 3 | Adjust for 25 % more hours per day. | New hours/day = ( 8 \times 1.25 = 10 ) hrs/day. | 10 hrs/day |
| 4 | Compute new daily productivity. | ( 18.75\text{ kg/hr} \times 10\text{ hrs} = 187.5\text{ kg/day} ) | 187.5 kg/day |
| 5 | Find total for next 5 days. | ( 187.5 \times 5 = 937.5\text{ kg} ) | 937.5 kg |
Answer: 937.5 kg of timber will be collected in the next 5 days.
Quick Check
- Productivity per hour unchanged → only hours affect total.
- 25 % increase ⇒ 1.25 × 8 = 10 hrs → 25 % increase in daily yield.
- 187.5 kg/day × 5 = 937.5 kg. ✔️
4. Quick Revision Summary
| Topic | Formula | Quick Hint |
|---|---|---|
| Ratio | ( \frac{a}{b} = \frac{c}{d} ) | “Cross‑multiply to avoid fractions.” |
| Percentage | ( P% \text{ of } X = \frac{P}{100}X ) | Convert percentage to decimal first. |
| Time‑Work | Work = Rate × Time | Rate = Work/Time; keep “units per hour/day”. |
| Distance | ( d = r \times t ) | Use |